A Novel Look at the One Dimensional Delta Schrodinger Equation By Patrick Bruskiewich Department of Physics and Astronomy, University of British Columbia email: patrickb@physics . ubc . ca Resume In this paper both integral transforms and an algebraic reformulation of the second order differential equation are used to derive the solution to the one dimensional attractive delta Schrodinger equation. This is a novel approach not found in any standard textbook. Both the transform and the reformulation methods provide additional insight into the nature of the one dimensional attractive delta potential. Introduction In many problems in quantum mechanics one can arrive at a solution in several ways. It is sometimes interesting to revisit a familar problem to see what new can be learned. Both algebraic and analytic techniques are used in this paper to solve for the one dimensional Schrodinger equation with an attractive delta potential. An attractive delta potential can be used to model a point particle, one dimensional models of molecules with multiple centres and in the Kronig- Penney model of crystalline structure to name just a few occurences of the potential in the literature. It is also of central importance in the Green's Function formulation of quantum mechanics. 22 Gamma 138 Patrick Bruskiewich The traditional manner in which to work with the one dimensional Schrodinger equation is through differentiation. The fundamental equa- tions of quantum mechanics can also be derived from a variational prin- ciple or by means of integration. This has direct bearing on the manner in which the solution to the one dimensional attractive delta Schrodinger equation is derived in this paper. We must begin by being mindful that the delta function is not a func- tion in the traditional sense. It is a generalized function, or distribution, which can be thought of as a limit of a sequence of functions of ever increasing height and ever-decreasing width, used as an integrand in an integral equation. For example in its simplest representation, we can model an attractive delta potential by a square well potential in the limit that its depth (Vo) approaches infinity while its width (a) approaches but never reaches zero in the limit, (with the proviso that Vo a ~ 1 in the limit). In the limit we can also represent a delta function in several other relevant ways, namely s{x) = 7?^7? exp[ -7 5(x) = — lim 7T x->0 x 2 + e 2 _, . 1 siniVa: dlx) = — lim As is evident, each of these representations has a characteristic width which approaches, but never reaches zero in their limit. This is of impor- tance when we remember that a physical state must always correspond to a square-integrable wavefunction. If 6(x — a) represents a delta function at point a then given some function f(x) 23 A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 /OO rOO f(x)5(x — a)dx = fia) / 6(x — a)dx = fid) -00 ' J — OO The standard interpretation of the delta function, irrespective of its functional representation, is that under the integral sign the delta func- tion serves to "pick out" the value of the function fix) at the point a. The Traditional Derivation Consider the one dimensional Schrodinger equation with an attractive delta potential given by H 2 d 2 ip(x) — a 6(x) ip(x) = E ift(x). 2m dx 2 The standard way to solve this equation is to match continuity condi- tions. For a potential centred on x = that spans — oo < x < +oo we consider both the Dirichlet boundary condition and the Neumann bound- ary condition. [1] Let us consider the representation of the delta function as a box of small width 2e centred on x = and large negative potential —a. In the region x < — e, the potential is zero so we have a solution of the form ift-(x) = Aex.p(nx) where k = ^ — ^ . Similarly in the region x > e the potential is again zero and in this region we have a solution of the form ip+(x) = .Bexp(— kx) 24 Gamma 138 Patrick Bruskiewich It remains only to match these two functions at x = with the bound- dx ary conditions, namely ip is continuous and -f- is continuous except at points where the potential is infinite. Evidently the delta function must describe the discontinuity in the derivative of ip at x = 0. We can extract the delta function by integrating the Schrodinger equation, from — e to +e, and then take the limit as e — > 0, namely / !T~2 ^( x ) dx — a 5(x) ip(x) dx = E ^{x) dx. In the limit the last integral is clearly equal to zero, while the first integral is /: S «•> - - *2> which is a measure as to how much the slope in the wavefunction changes over the interval — e < x < e. Thus we have simplified the expression to .,dib. —2ma .. r+e —2ma , , n . A( i ) = -&- H L *« *w dx = n^ W) Through integration we have reduced the order of the Schrodinger equation from second to first order and we incorporated the boundary conditions in the first order equation explicitly. By symmetry we recognize that Solving this first order differential equation we arrive at the familiar normalized wavefunction solution given by 25 A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 , . y/2ma . ma I x I * (x) = -jr~ exp( — w~ ] The energy of this normalized wavefunction solution is given by ma 2 £ = ~w There are several other ways in which to derive the solution to the one dimensional Schrodinger equation with an attractive delta potential. The alternate techniques outlined in this paper can act as a starting point to describing the quantum behaviour of other systems. Solution by an Integral Transform Consider once again the one dimensional attractive delta Schrodinger equation h 2 d 2 ip — a 5(x) tjj = E tp. 2m dx 2 We can borrow a technique from quantum field theory and derive a so- lution using an integral transform in the way of a Fourier transform. [2] Let 1 °r ^(x) = — dk e lkx cf>(k) — 00 where <j){k) is the wave function in the momentum representation. Intro- ducing this solution into the differential equation we find that (^ k 2 + B) m = a V>(0) 26 Gamma 138 Patrick Bruskiewich where "0(0) is the wavefunction value at x = and B of the eigensolution. Solving for <j){k) we find E, is the energy m a *(0) B 2 + B Using the completeness theorem, 00 00 00 1 = I $*{x) if)(x) dx = J J dk dk' J(k-k/)x ±*( h i — 00 — 00 —00 2tt 2tt WW*) which when integrated over k' simplifies to -i 00 — / dk (f)*(k) <f>(k) — 00 2maip(ti) h 2 -i 00 - /- 7T i dk [fc 2 + 2mB ] 2 where we recognize that the integrand is an even function and change the integral limits accordingly. Upon integration we arrive at an inverse tangent function, namely tf 2maijj(0)_ 2 7T n i 3 r kh - k=0 ° arctan V2mB V2mB k=0 which when we solve for E yields (E = — B) E = - ma 2h 2 We can use our value for B in the wavefunction (j){k) and solve for ip(x) in the transform equation, namely i/)(x) = 2ma V>(0) h 2 °r dk -Akx -oo 2tt k 2 +[ ti 2 27 A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 By inspection we see that only the even function cos(A;a:) yields a non- zero contribution, and so adjusting the limits of integration accordingly we find *(*) = — si— J- 2ma "0(0) °f dk cos(kx) k 2 + [f^ 12 which yields the solution il){x) = V>(0)exp( — — ) Solution by Reformulation We can find a solution to the one dimensional attractive delta Schrodinger equation by using a simple transformation to the second order differential equation. [3] Consider the delta function S(x) represented by the second partial derivative of the absolute value function in x, namely 1 d 2 *(*) = 2W I X I • Gathering the terms in the second derivative we can rewrite the Schrodinger equation as h " d " ! a r Q2 i 11/ w , 2m dx 2 2 dx 2 Bringing the wavefunction ip over to the left hand side, we get 1 d ,dib s ma d ,d\x\ s 2m I E ip dx dx h 2 dx dx h 2 28 Gamma 138 Patrick Bruskiewich We have used the fact that for a bounded wavefunction the energy E is less than zero, namely E = -\E\. Bringing the wavefunction ip into the bracketed derivative and gather- ing terms we have d , 1 dip ma d I x L 1 , dip . 9 2ra I E I — ( — H ! — -) H (— -) = ! -■ dx ip dx H 2 dx ip 2 dx h 2 Let us set the following condition on tp d A dip ma d \ x \ dx ip dx h 2 dx Solving this equation we get ^ = ■0(0) exp( — — | x |). Using the normalization condition we find the familiar wavefunction for the delta potential . J ma , ma . .. ^ = __ exp( __ | x p. Having found ip we can now use this wavefunction to solve for the remaining condition, namely the value for the energy of the wavefunction, 1 M.n 2m I E "(-ST) = ip 2 dx h 2 Solving for the Energy (| E |= — E we get „ ma 2 E = 2h 2 ' which is the expected value for the energy of a bounded wavefunction for the delta potential. 29 A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 Solution by a Laplace Transform Consider once again the one dimensional attractive delta Schrodinger equation 2 £>2 K l d 2m dx 2 ip — a 5(x) ip = E i\). We can solve for this second order differential equation using the Laplace transform. Applying the usual transform 00 L(f) = Jf(t)e- st dt to this second order differential equation yields ( s 2 + -ft- ) L (VO + ( -^- - s) V^(O) - <//(0) = Let i/j(0) = C and ^'(0) = 7 so that 9 2mE 2ma " ' ) L (VO + ( -r^- - s ) C - 7 (* 2 + ft 2 ft 2 Solving for L (*0) provides the following transform equation which needs to be solved: MVO (7-^0 ( s 2 + b 2 ) ' ( s 2 + b 2 ) *c + where 6 2 = -^-. The inverse transform of the first term is L- 1 (s 2 + b 2 ) ( cos ibx) 30 Gamma 138 Patrick Bruskiewich If we let r\ = (7 — ^p C) then the inverse transform of the second term is of the form r\ L -1 2ib (s-ib) 2ib (s + ib) V 2ib kl Jbx —ibx ] 1 b sin (bx) So the solution for a one dimensional attractive delta Schrodinger equa- tion using the Laplace transform method is tjj(x) Tj C cos (bx) + — sin (bx) ( cos (bx) + 1 7 2ma c sin (bx) which appears of a very different form to that of the solution given in part one of this paper. However, it is possible to rewrite this eigensolution in a more familiar form. If we have a bounded solution we can change E — > E I giving ip(x) = £ cosh (| b I x) — = C cosh (| b I x) + 77 sinh (| b I x) 7 2ma c sinh (| b I x) since b -> i \ b |, cos (bx) — > cosh (| b \ x) and sin (bx) — >• i sinh (| 6 | #) under this tranformation. Since we are in the domain < x < 00 in our solution we need to reformulate that part of cosh (bx) to conform to our boundary conditions as # — > 00. We can do this by recognizing that C exp (— I b I x) = ( [ cosh (| b \ x) — sinh (| b \ x) m \a\ At this point, analysis of the sinh (bx) term shows that \ b \ n , that this term can be rewritten as so 31 A Novel Look at the One Dimensional Delta Schrodinger Equation Gamma 138 1 1 b\ 7 - 2ma h 2 C J 7 - ma h 2 C J sinh (| sinh (| b | x) = sinh (| b | x) + (, sinh (| b \ x) We see then we have tp(x) = £ exp (— | b | x) + ma c 7 sinh (| b | x) where the first term is the familiar solution to the one dimensional at- tractive delta Schrodinger equation. There is still the second term, the sinh (| b | x) part of the solution, we need to interpret. Recalling that ift'(0) = 7 we see that the term in sinh(| b \ x) is, in fact 1 T ma C - 7 sinh (| b I x) = c - <//(0) sinh (| b I x) By inspection we know that as x — > 00, the function sinh(| b \ x) — > 00. With some thought, however, and a careful consideration of the sign, in light of the | b \ term in the expression c V/(0) one sees immediately that we recover the well known solution to the one dimensional Schrodinger equation with an attractive delta potential, namely , . y/2ma , ma I x L tp{x) = — r — exp( — ) h h 2 32 Gamma 138 Patrick Bruskiewich Discussion The algebraic reformulation outlined above is a novel approach not found in the literature, as is the Laplace Transform approach. The use of the Laplace transform to solve one dimensional equations in quantum me- chanics is an under-utilized technique which should be explored fur- ther. The direct mathematical tie between the Laplace transform and the Fourier transform method is not so obvious, but is rich in implication. Along with having applicability in quantum mechanics, the one dimen- sional attractive delta potential is a good starting point to the study of the Dirac delta potential in more than one dimension. This multidimen- sional potential in turn allows the introduction within the framework of quantum mechanics of many of the basic concepts of quantum field the- ory, such as regularization, the renormalization group and dimensional transformation. References [1] The standard derivation of the solution to the one dimensional attrac- tive delta Schrodinger equation can be found in most any quantum textbook, such as the texts by Dirac, Messiah, Merzbacher, Grif- fiths or Gasiorowicz. [2] For a succinct overview of this technique refer to section 10.8 of the book Quarks, Leptons and Gauge Fields by Kerson Huang, World Scientific, Singapore, 1982. [3] The idea for this approach derives from a comment in section 6.2 of Quantum Mechanics, 2nd Edition by Eugen Merzbacher, John Wiley and Sons, New York, 1970. 33